Problem: Simplify the following expression and state the conditions under which the simplification is valid. You can assume that $n \neq 0$. $z = \dfrac{n - 5}{-4n + 8} \div \dfrac{-6n + 30}{n^2 - 11n + 18} $
Dividing by an expression is the same as multiplying by its inverse. $z = \dfrac{n - 5}{-4n + 8} \times \dfrac{n^2 - 11n + 18}{-6n + 30} $ First factor the quadratic. $z = \dfrac{n - 5}{-4n + 8} \times \dfrac{(n - 2)(n - 9)}{-6n + 30} $ Then factor out any other terms. $z = \dfrac{n - 5}{-4(n - 2)} \times \dfrac{(n - 2)(n - 9)}{-6(n - 5)} $ Then multiply the two numerators and multiply the two denominators. $z = \dfrac{ (n - 5) \times (n - 2)(n - 9) } { -4(n - 2) \times -6(n - 5) } $ $z = \dfrac{ (n - 5)(n - 2)(n - 9)}{ 24(n - 2)(n - 5)} $ Notice that $(n - 5)$ and $(n - 2)$ appear in both the numerator and denominator so we can cancel them. $z = \dfrac{ \cancel{(n - 5)}(n - 2)(n - 9)}{ 24\cancel{(n - 2)}(n - 5)} $ We are dividing by $n - 2$ , so $n - 2 \neq 0$ Therefore, $n \neq 2$ $z = \dfrac{ \cancel{(n - 5)}\cancel{(n - 2)}(n - 9)}{ 24\cancel{(n - 2)}\cancel{(n - 5)}} $ We are dividing by $n - 5$ , so $n - 5 \neq 0$ Therefore, $n \neq 5$ $z = \dfrac{n - 9}{24} ; \space n \neq 2 ; \space n \neq 5 $